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Sum of binomial coefficients and cos(kx)

  1. Jul 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculate the following sum:
    ScreenShot035.jpg (click to expand)


    3. The attempt at a solution
    I tried something with Moivre formula and Newton binomial theorem but no result :redface:, should i continue with these or is there any simpler approach?

    I just need some hints.

    Thanks.
     
  2. jcsd
  3. Jul 6, 2011 #2

    Ray Vickson

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    cos(k*x) = (1/2)[exp(i*k*x) + exp(-i*k*x)], where i = sqrt(-1), and exp(i*k*x) = z^k for an appropriate z.

    RGV
     
  4. Jul 6, 2011 #3
    Thanks for the hint but i cannot still work it out.

    I plugged cos(kx) directly in the sum and resulted in some exponentials..

    [URL]http://www2.wolframalpha.com/Calculate/MSP/MSP256719g79b7bde68e6h6000020i46ai95ba26aii?MSPStoreType=image/gif&s=20&w=225&h=36[/URL]

    The result should look like this : [URL]http://www2.wolframalpha.com/Calculate/MSP/MSP126419g79d95ifaec8fe00001ifd1gdagh48hf7e?MSPStoreType=image/gif&s=3&w=152&h=36[/URL]

    May i get one more hint ?:p
     
    Last edited by a moderator: Apr 26, 2017
  5. Jul 6, 2011 #4

    SammyS

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    Try a multiple angle formula for cos(nx) such as:
    [tex]\cos(n x) = \cos\left({{n x}\over{3}}\right) \left(-1+2 \cos\left({{2 (n x)}\over{3}}\right)\right)[/tex]
    or
    [tex]\cos(n x) = -1+2\cos^2\left({{n x}\over{2}}\right)[/tex]
     
  6. Jul 6, 2011 #5

    vela

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    It's easier to use the fact that cos nx = real part of einx. The idea is to calculate a related sum where the real part is the answer you're looking for.
     
  7. Jul 6, 2011 #6
    Ok, thank you both for the ideas, i'll try to solve it tomorrow and come back with a result.
     
  8. Jul 7, 2011 #7
    I wasn't able to come up with a result.

    Here is my trial:

    i got this result using euler's formula
    (e^(ix)+1)^n + (e^(-ix)+1)^n = 2 SUM(k=0,n) C{k,n} * cos(k*x)

    I tried how @vela said but i couldn't figure out what sum to calculate.
     
  9. Jul 7, 2011 #8

    Ray Vickson

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    I don't understand your comment. You have already computed the sum, so you already have the answer. However, maybe the person who posed the problem does not want to see complex quantities, and in that case you need to extract the real part of (1+exp(ix))^n, which you can do by converting 1 + exp(ix) to polar form, for example.

    RGV
     
  10. Jul 7, 2011 #9

    vela

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    Use the fact that
    \begin{eqnarray*}
    e^{ix}+1 &= e^{ix/2}(e^{ix/2}+e^{-ix/2}) \\
    e^{-ix}+1 &= e^{-ix/2}(e^{ix/2}+e^{-ix/2})
    \end{eqnarray*}
    Using my suggestion would have given you
    [tex]\sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} \cos kx = \sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} \mathrm{Re}(e^{ikx}) = \mathrm{Re}\left[\sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} e^{ikx}\right] = \cdots[/tex]
     
  11. Jul 7, 2011 #10
    Well this is too much :), i'm lost in the imaginary world, i'll give it a try another time.
    Your help is appreciated, thank you.
     
  12. Jul 9, 2011 #11

    vela

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    You'll find Euler's formula
    [tex]e^{i\theta} = \cos\theta + i\sin\theta[/tex]
    very helpful in general. Using it, you can express the trig functions in terms of complex exponentials
    \begin{align*}
    \cos\theta &= \frac{e^{i\theta}+e^{-i\theta}}{2} \\
    \sin\theta &= \frac{e^{i\theta}-e^{-i\theta}}{2i}
    \end{align*}
    Back to your problem... You don't have much left to do to get the answer. For example, you have
    [tex](e^{ix}+1)^n = [e^{ix/2}(e^{ix/2}+e^{-ix/2})]^n = e^{inx/2}(e^{ix/2}+e^{-ix/2})^n[/tex]
    Nothing terribly complicated happening there, just algebra. Now if you compare what's inside the parentheses to the formulas above, you can see or show it's equal to
    [tex]e^{inx/2}(e^{ix/2}+e^{-ix/2})^n = e^{inx/2}[2\cos(x/2)]^n[/tex]
    Now simplify the term [itex](e^{-ix}+1)^n[/itex] similarly and then combine the results.
     
  13. Jul 10, 2011 #12
    Thank you for the explanation!
     
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